Question: 4] c) Find the roots of the equation $x^4+x^3-7x^2-x+5=0$ which lies between 2 and 2.1 correct to 3 places of decimals using Regula Falsi method.
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Subject:- Applied Mathematics

Marks:- 3

Mumbai Unversity>FE>Sem1>Applied Maths1

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modified 8 weeks ago  • written 8 weeks ago by gravatar for Mayank Aggarwal Mayank Aggarwal0
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Given that a=2 and b =2.1.

$f(2) = (2)^4+(2)^3-7(2)^2-2+5= -1$.

$f(2.1) = (2.1)^4+(2.1)^3-7(2.1)^2-(2.1)+5 = 0.739100.$

$x_1 = \frac{(af(b)-bf(a))}{(f(b)-f(a) )} = (2×0.73910—1))×(2.1)0.739100—10.739100—1.05750. ……………(1)$

$f((x_1 ) = (2.05750)^4+(2.05750)^3-7(2.05750)^2-2.05750+5 = -0.05973.$

$x_2 = \frac{(af(x_1 )- x_1 f(a))}{(f(x_1 )-f(a) )} = (2×(-0.05973)—1))×(2.05750)0.05973—1.05750)0.05973—12 ……………(2)$

$f((x_2 ) = (2.061152)^4+(2.061152)^3-7(2.061152)^2-2.061152+5 = 0.005326.$

$x_3 = \frac{(af(x_2 )- x_2 f(a))}{(f(x_2 )-f(a) )} = (2×(0.005326)—1))×(2.061152)0.005326—1.061152)0.005326—1. ………………(3)$

$f((x_3 ) = (2.06082)^4+(2.06082)^3-7(2.06082)^2-2.06082+5 = -0.000582.$

$x_4 = \frac{(af(x_3 )- x_3 f(a))}{(f(x_3 )-f(a) )} = (2×(-0.000582)—1))×(2.06082)0.000582—12.06082)0.000582—1 ………………(4)$

Hence from (4) and (3) iteration we get that value of x is coinciding.

Therefore the final value of x is 2.0608.

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written 8 weeks ago by gravatar for Mayank Aggarwal Mayank Aggarwal0
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